Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
The answer is 500.902 hope you get it right
Answer:
if your asking what do you have to add to negative 4 to get 16then you add 20 or if your asking what to add to get -4 I will need more info message me or comment
Answer:
m > 5 or m < -19
Step-by-step explanation:
We are given the Inequality:

First, add both sides by 4.

<u>A</u><u>b</u><u>s</u><u>o</u><u>l</u><u>u</u><u>t</u><u>e</u><u> </u><u>V</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>P</u><u>r</u><u>o</u><u>p</u><u>e</u><u>r</u><u>t</u><u>y</u>

Given a = any expressions and b = any positive numbers, zero or any expressions.

From the Inequality, change > to equal

Subtract 7 both sides.

±12-7 can be 12-7 = 5 or -12-7 = -19

Refer to the attachment. The region is when the absolute value function is greater than constant function y = 12 or a blue horizontal line.
Since the absolute graph is above constant graph when x > 5 and x <-19.
Therefore,

2.4,3, 7 1/3, 221/30,7.36 is order from least to greatest